How to reverse a singly linked list using only two pointers?

I hate to be the bearer of bad news but I don't think your three-pointer solution actually works. When I used it in the following test harness, the list was reduced to one node, as per the following output:

==========
4
3
2
1
0
==========
4
==========

You won't get better time complexity than your solution since it's O(n) and you have to visit every node to change the pointers, but you can do a solution with only two extra pointers quite easily, as shown in the following code:

#include <stdio.h>

// The list element type and head.

struct node { 
    int data;
    struct node *link;
};
static struct node *first = NULL;

// A reverse function which uses only two extra pointers.

void reverse() {
    // curNode traverses the list, first is reset to empty list.
    struct node *curNode = first;
    first = NULL;

    // Until no more in list, insert current before first and advance.
    while (curNode != NULL) {
        // Need to save next node since we're changing the current.
        struct node *nxtNode = curNode->link;

        // Insert at start of new list.
        curNode->link = first;
        first = curNode;

        // Advance to next.
        curNode = nxtNode;
    }
}

// Code to dump the current list.

static void dumpNodes() {
    struct node *curNode = first;
    printf ("==========\n");
    while (curNode != NULL) {
        printf ("%d\n", curNode->data);
        curNode = curNode->link;
    }
}

// Test harness main program.

int main (void) {
    int i;
    struct node *newnode;

    // Create list (using actually the same insert-before-first
    // that is used in reverse function.

    for (i = 0; i < 5; i++) {
        newnode = malloc (sizeof (struct node));
        newnode->data = i;
        newnode->link = first;
        first = newnode;
    }

    // Dump list, reverse it, then dump again.

    dumpNodes();
    reverse();
    dumpNodes();
    printf ("==========\n");

    return 0;
}

This code outputs:

==========
4
3
2
1
0
==========
0
1
2
3
4
==========

which I think is what you were after. It can actually do this since, once you've loaded up first into the pointer traversing the list, you can re-use first at will.

Any alternative? No, this is as simple as it gets, and there's no fundamentally-different way of doing it. This algorithm is already O(n) time, and you can't get any faster than that, as you must modify every node.

It looks like your code is on the right track, but it's not quite working in the form above. Here's a working version:

#include <stdio.h>

typedef struct Node {
  char data;
  struct Node* next;
} Node;

void print_list(Node* root) {
  while (root) {
    printf("%c ", root->data);
    root = root->next;
  }
  printf("\n");
}

Node* reverse(Node* root) {
  Node* new_root = 0;
  while (root) {
    Node* next = root->next;
    root->next = new_root;
    new_root = root;
    root = next;
  }
  return new_root;
}

int main() {
  Node d = { 'd', 0 };
  Node c = { 'c', &d };
  Node b = { 'b', &c };
  Node a = { 'a', &b };

  Node* root = &a;
  print_list(root);
  root = reverse(root);
  print_list(root);

  return 0;
}

#include <stddef.h>

typedef struct Node {
    struct Node *next;
    int data;
} Node;

Node * reverse(Node *cur) {
    Node *prev = NULL;
    while (cur) {
        Node *temp = cur;
        cur = cur->next; // advance cur
        temp->next = prev;
        prev = temp; // advance prev
    }
    return prev;
}